I've created a simple Spring-Application with a MySQL-DB.
In the DB there are 20 years of stock data (5694 lines):
The Goal is to find the Best Moving Average (N) for that 20 years of stock data. Inputs are the closing prices of every trading day.
The calculated average depends on the N. So p.e. if N=3 the average of a reference day t, is given by ((t-1)+(t-2)+(t-3)/N).
Output is the Best Moving Average (N) and the Result you made with all the buying & selling transactions of the Best N.
I did not find a proper algorithm in the Internet, so I implemented the following:
For every N (249-times) the program does the following steps:
SQL-Query: calculates averages & return list
#Repository
public interface StockRepository extends CrudRepository<Stock, Integer> {
/*
* This sql query calculate the moving average of the value n
*/
#Query(value = "SELECT a.date, a.close, Round( ( SELECT SUM(b.close) / COUNT(b.close) FROM stock AS b WHERE DATEDIFF(a.date, b.date) BETWEEN 0 AND ?1 ), 2 ) AS 'avg' FROM stock AS a ORDER BY a.date", nativeQuery = true)
List<AverageDTO> calculateAverage(int n);
Simulate buyings & sellings – > calculate result
Compare result with bestResult
Next N
#RestController
public class ApiController {
#Autowired
private StockRepository stockRepository;
#CrossOrigin(origins = "*")
#GetMapping("/getBestValue")
/*
* This function tries all possible values in the interval [min,max], calculate
* the moving avg and simulate the gains for each value to choose the best one
*/
public ResultDTO getBestValue(#PathParam("min") int min, #PathParam("max") int max) {
Double best = 0.0;
int value = 0;
for (int i = min; i <= max; i++) {
Double result = simulate(stockRepository.calculateAverage(i));
if (result > best) {
value = i;
best = result;
}
}
return new ResultDTO(value, best);
}
/*
* This function get as input the close and moving average of a stock and
* simulate the buying/selling process
*/
public Double simulate(List<AverageDTO> list) {
Double result = 0.0;
Double lastPrice = list.get(0).getClose();
for (int i = 1; i < list.size(); i++) {
if (list.get(i - 1).getClose() < list.get(i - 1).getAvg()
&& list.get(i).getClose() > list.get(i).getAvg()) {
// buy
lastPrice = list.get(i).getClose();
} else if (list.get(i - 1).getClose() > list.get(i - 1).getAvg()
&& list.get(i).getClose() < list.get(i).getAvg()) {
// sell
result += (list.get(i).getClose() - lastPrice);
lastPrice = list.get(i).getClose();
}
}
return result;
}
}
When I put Min=2 and Max=250 it takes 45 minutes to finish.
Since, I'm a beginner in Java & Spring I do not know how I can optimize it.
I'm happy for every input.
This problem is equivalent with finding the best moving N sum. Simply then divide by N. Having such a slice, then the next slice subtracts the first value and adds a new value to the end. This could lead to an algorithm for finding local growths with a[i + N] - a[i] >= 0.
However in this case a simple sequential ordered query with
double[] slice = new double[N];
double sum = 0.0;
suffices. (A skipping algorithm on a database is probably too complicated.)
Simply walk through the table keeping the slice as window, keeping N values and keys, and maintaining the maximum upto now.
Use the primitive type double instead of the object wrapper Double.
If the database transport is a serious factor, a stored procedure would do. Keeping a massive table as really many entities just for a running maximum is unfortunate.
It would be better to have a condensed table or better field with the sum of N values.
Related
I have a program that takes in anywhere from 20,000 to 500,000 velocity vectors and must output these vectors multiplied by some scalar. The program allows the user to set a variable accuracy, which is basically just how many decimal places to truncate to in the calculations. The program is quite slow at the moment, and I discovered that it's not because of multiplying a lot of numbers, it's because of the method I'm using to truncate floating point values.
I've already looked at several solutions on here for truncating decimals, like this one, and they mostly recommend DecimalFormat. This works great for formatting decimals once or twice to print nice user output, but is far too slow for hundreds of thousands of truncations that need to happen in a few seconds.
What is the most efficient way to truncate a floating-point value to n number of places, keeping execution time at utmost priority? I do not care whatsoever about resource usage, convention, or use of external libraries. Just whatever gets the job done the fastest.
EDIT: Sorry, I guess I should have been more clear. Here's a very simplified version of what I'm trying to illustrate:
import java.util.*;
import java.lang.*;
import java.text.DecimalFormat;
import java.math.RoundingMode;
public class MyClass {
static class Vector{
float x, y, z;
#Override
public String toString(){
return "[" + x + ", " + y + ", " + z + "]";
}
}
public static ArrayList<Vector> generateRandomVecs(){
ArrayList<Vector> vecs = new ArrayList<>();
Random rand = new Random();
for(int i = 0; i < 500000; i++){
Vector v = new Vector();
v.x = rand.nextFloat() * 10;
v.y = rand.nextFloat() * 10;
v.z = rand.nextFloat() * 10;
vecs.add(v);
}
return vecs;
}
public static void main(String args[]) {
int precision = 2;
float scalarToMultiplyBy = 4.0f;
ArrayList<Vector> velocities = generateRandomVecs();
System.out.println("First 10 raw vectors:");
for(int i = 0; i < 10; i++){
System.out.print(velocities.get(i) + " ");
}
/*
This is the code that I am concerned about
*/
DecimalFormat df = new DecimalFormat("##.##");
df.setRoundingMode(RoundingMode.DOWN);
long start = System.currentTimeMillis();
for(Vector v : velocities){
/* Highly inefficient way of truncating*/
v.x = Float.parseFloat(df.format(v.x * scalarToMultiplyBy));
v.y = Float.parseFloat(df.format(v.y * scalarToMultiplyBy));
v.z = Float.parseFloat(df.format(v.z * scalarToMultiplyBy));
}
long finish = System.currentTimeMillis();
long timeElapsed = finish - start;
System.out.println();
System.out.println("Runtime: " + timeElapsed + " ms");
System.out.println("First 10 multiplied and truncated vectors:");
for(int i = 0; i < 10; i++){
System.out.print(velocities.get(i) + " ");
}
}
}
The reason it is very important to do this is because a different part of the program will store trigonometric values in a lookup table. The lookup table will be generated to n places beforehand, so any velocity vector that has a float value to 7 places (i.e. 5.2387471) must be truncated to n places before lookup. Truncation is needed instead of rounding because in the context of this program, it is OK if a vector is slightly less than its true value, but not greater.
Lookup table for 2 decimal places:
...
8.03 -> -0.17511085919
8.04 -> -0.18494742685
8.05 -> -0.19476549993
8.06 -> -0.20456409661
8.07 -> -0.21434223706
...
Say I wanted to look up the cosines of each element in the vector {8.040844, 8.05813164, 8.065688} in the table above. Obviously, I can't look up these values directly, but I can look up {8.04, 8.05, 8.06} in the table.
What I need is a very fast method to go from {8.040844, 8.05813164, 8.065688} to {8.04, 8.05, 8.06}
The fastest way, which will introduce rounding error, is going to be to multiply by 10^n, call Math.rint, and to divide by 10^n.
That's...not really all that helpful, though, considering the introduced error, and -- more importantly -- that it doesn't actually buy anything. Why drop decimal points if it doesn't improve efficiency or anything? If it's about making the values shorter for display or the like, truncate then, but until then, your program will run as fast as possible if you just use full float precision.
I have an ArrayList of colors and their frequency of appearance. My program should calculate a reordering of those items that maximizes the minimum distance between two equal bricks.
For example, given input consisting of 4*brick 1 (x), 3*brick 2 (y), and 5*brick 3 (z), one correct output would be: z y x z x z y x z x y.
My code does not produce good solutions. In particular, sometimes there are 2 equal bricks at the end, which is the worst case.
import java.util.ArrayList;
import java.util.Collections;
public class Calc {
// private ArrayList<Wimpel> w = new ArrayList<Brick>();
private String bKette = "";
public String bestOrder(ArrayList<Brick> w) {
while (!w.isEmpty()) {
if (w.get(0).getFrequency() > 0) {
bChain += w.get(0).getColor() + "|";
Brick brick = new Wimpel(w.get(0).getVariant(), w.get(0).getFrequency() - 1);
w.remove(0);
w.add(brick);
// bestOrder(w);
} else {
w.remove(0);
}
bestOrder(w);
}
return bOrder;
}
public int Solutions(ArrayList<Wimpel> w) {
ArrayList<Brick> tmp = new ArrayList<Brick>(w);
int l = 1;
int counter = (int) w.stream().filter(c -> Collections.max(tmp).getFrequency() == c.getFrequency()).count();
l = (int) (fakultaet(counter) * fakultaet((tmp.size() - counter)));
return l;
}
public static long fakultaet(int n) {
return n == 0 ? 1 : n * fakultaet(n - 1);
}
}
How can make my code choose an optimal order?
We will not perform your exercise for you, but we will give you some advice.
Consider your current approach: it operates by filling the result string by cycling through the bricks, choosing one item from each brick in turn as long as any items remain in that brick. But this approach is certain to fail when one brick contains at least two items more than any other, because then only that brick remains at the end, and all its remaining items have to be inserted one after the other.
That is, the problem is not that your code is buggy per se, but rather that your whole strategy is incorrect for the problem. You need something different.
Now, consider the problem itself. Which items will appear at the shortest distance apart in a correct ordering? Those having the highest frequency, of course. And you can compute that minimum distance based on the frequency and total number of items.
Suppose you arrange these most-constrained items first, at the known best distance.
What's left to do at this point? Well, you potentially have some more bricks with lesser frequency, and some more slots in which to accommodate their items. If you ignore the occupied slots altogether, you can treat this as a smaller version of the same problem you had before.
I am trying to build an OCR by calculating the Coefficient Correlation between characters extracted from an image with every character I have pre-stored in a database. My implementation is based on Java and pre-stored characters are loaded into an ArrayList upon the beginning of the application, i.e.
ArrayList<byte []> storedCharacters, extractedCharacters;
storedCharacters = load_all_characters_from_database();
extractedCharacters = extract_characters_from_image();
// Calculate the coefficent between every extracted character
// and every character in database.
double maxCorr = -1;
for(byte [] extractedCharacter : extractedCharacters)
for(byte [] storedCharacter : storedCharactes)
{
corr = findCorrelation(extractedCharacter, storedCharacter)
if (corr > maxCorr)
maxCorr = corr;
}
...
...
public double findCorrelation(byte [] extractedCharacter, byte [] storedCharacter)
{
double mag1, mag2, corr = 0;
for(int i=0; i < extractedCharacter.length; i++)
{
mag1 += extractedCharacter[i] * extractedCharacter[i];
mag2 += storedCharacter[i] * storedCharacter[i];
corr += extractedCharacter[i] * storedCharacter[i];
} // for
corr /= Math.sqrt(mag1*mag2);
return corr;
}
The number of extractedCharacters are around 100-150 per image but the database has 15600 stored binary characters. Checking the coefficient correlation between every extracted character and every stored character has an impact on the performance as it needs around 15-20 seconds to complete for every image, with an Intel i5 CPU.
Is there a way to improve the speed of this program, or suggesting another path of building this bringing similar results. (The results produced by comparing every character with such a large dataset is quite good).
Thank you in advance
UPDATE 1
public static void run() {
ArrayList<byte []> storedCharacters, extractedCharacters;
storedCharacters = load_all_characters_from_database();
extractedCharacters = extract_characters_from_image();
// Calculate the coefficent between every extracted character
// and every character in database.
computeNorms(charComps, extractedCharacters);
double maxCorr = -1;
for(byte [] extractedCharacter : extractedCharacters)
for(byte [] storedCharacter : storedCharactes)
{
corr = findCorrelation(extractedCharacter, storedCharacter)
if (corr > maxCorr)
maxCorr = corr;
}
}
}
private static double[] storedNorms;
private static double[] extractedNorms;
// Correlation between to binary images
public static double findCorrelation(byte[] arr1, byte[] arr2, int strCharIndex, int extCharNo){
final int dotProduct = dotProduct(arr1, arr2);
final double corr = dotProduct * storedNorms[strCharIndex] * extractedNorms[extCharNo];
return corr;
}
public static void computeNorms(ArrayList<byte[]> storedCharacters, ArrayList<byte[]> extractedCharacters) {
storedNorms = computeInvNorms(storedCharacters);
extractedNorms = computeInvNorms(extractedCharacters);
}
private static double[] computeInvNorms(List<byte []> a) {
final double[] result = new double[a.size()];
for (int i=0; i < result.length; ++i)
result[i] = 1 / Math.sqrt(dotProduct(a.get(i), a.get(i)));
return result;
}
private static int dotProduct(byte[] arr1, byte[] arr2) {
int dotProduct = 0;
for(int i = 0; i< arr1.length; i++)
dotProduct += arr1[i] * arr2[i];
return dotProduct;
}
Nowadays, it's hard to find a CPU with a single core (even in mobiles). As the tasks are nicely separated, you can do it with a few lines only. So I'd go for it, though the gain is limited.
In case you really mean cross-correlation, then a transform like DFT or DCT could help. They surely do for big images, but with yours 12x16, I'm not sure.
Maybe you mean just a dot product? And maybe you should tell us?
Note that you actually don't need to compute the correlation, most of the time you only need is find out if it's bigger than a threshold:
corr = findCorrelation(extractedCharacter, storedCharacter)
..... more code to check if this is the best match ......
This may lead to some optimizations or not, depending on how the images look like.
Note also that a simple low level optimization can give you nearly a factor of 4 as in this question of mine. Maybe you really should tell us what you're doing?
UPDATE 1
I guess that due to the computation of three products in the loop, there's enough instruction level parallelism, so a manual loop unrolling like in my above question is not necessary.
However, I see that those three products get computed some 100 * 15600 times, while only one of them depends on both extractedCharacter and storedCharacter. So you can compute
100 + 15600 + 100 * 15600
dot products instead of
3 * 100 * 15600
This way you may get a factor of three pretty easily.
Or not. After this step there's a single sum computed in the relevant step and the problem linked above applies. And so does its solution (unrolling manually).
Factor 5.2
While byte[] is nicely compact, the computation involves extending them to ints, which costs some time as my benchmark shows. Converting the byte[]s to int[]s before all the correlations gets computed saves time. Even better is to make use of the fact that this conversion for storedCharacters can be done beforehand.
Manual loop unrolling twice helps but unrolling more doesn't.
I have a piece of code that must run extremely fast in terms of clock speed. The algorithm is already in O(N). It takes 2seconds, it needs to take 1s. For most A.length inputs ~ 100,000 it takes .3s unless a particular line of code is invoked an extreme number of times. (For an esoteric programming challenge)
It uses a calculation of the arithmetic series that 1,2,..N -> 1,3,4,10,15..
that can be represented by n*(n+1)/2
I loop through this equation hundreds of thousands of times.
I do not have access to the input, nor can I display it. The only information I am able to get returned is the time it took to run.
particularly the equation is:
s+=(n+c)-((n*(n+1))/2);
s and c can have values range from 0 to 1Billion
n can range 0 to 100,000
What is the most efficient way to write this statement in terms of clock speed?
I have heard division takes more time then multiplication, but beyond that I could not determine whether writing this in one line or multiple assignment lines was more efficient.
Dividing and multiplying versus multiplying and then dividing?
Also would creating custom integers types significantly help?
Edit as per request, full code with small input case (sorry if it's ugly, I've just kept stripping it down):
public static void main(String[] args) {
int A[]={3,4,8,5,1,4,6,8,7,2,2,4};//output 44
int K=6;
//long start = System.currentTimeMillis();;
//for(int i=0;i<100000;i++){
System.out.println(mezmeriz4r(A,K));
//}
//long end = System.currentTimeMillis();;
// System.out.println((end - start) + " ms");
}
public static int mezmeriz4r(int[]A,int K){
int s=0;
int ml=s;
int mxl=s;
int sz=1;
int t=s;
int c=sz;
int lol=50000;
int end=A.length;
for(int i=sz;i<end;i++){
if(A[i]>A[mxl]){
mxl=i;
}else if(A[i]<A[ml]){
ml=i;
}
if(Math.abs(A[ml]-A[mxl])<=K){
sz++;
if(sz>=lol)return 1000000000;
if(sz>1){
c+=sz;
}
}else{
if(A[ml]!=A[i]){
t=i-ml;
s+=(t+c)-((t*(t+1))/(short)2);
i=ml;
ml++;
mxl=ml;
}else{
t=i-mxl;
s+=(t+c)-((t*(t+1))/(short)2);
i=mxl;
mxl++;
ml=mxl;
}
c=1;
sz=0;
}
}
if(s>1000000000)return 1000000000;
return s+c;
}
Returned from Challenge:
Detected time complexity:
O(N)
test time result
example
example test 0.290 s. OK
single
single element 0.290 s. OK
double
two elements 0.290 s. OK
small_functional
small functional tests 0.280 s. OK
small_random
small random sequences length = ~100 0.300 s. OK
small_random2
small random sequences length = ~100 0.300 s. OK
medium_random
chaotic medium sequences length = ~3,000 0.290 s. OK
large_range
large range test, length = ~100,000 2.200 s. TIMEOUT ERROR
running time: >2.20 sec., time limit: 1.02 sec.
large_random
random large sequences length = ~100,000 0.310 s. OK
large_answer
test with large answer 0.320 s. OK
large_extreme
all maximal value = ~100,000 0.340 s. OK
With a little algebra, you can simply the expression (n+c)-((n*(n+1))/2) to c-((n*(n-1))/2) to remove an addition operation. Then you can replace the division by 2 with a bit-shift to the right by 1, which is faster than division. Try replacing
s+=(n+c)-((n*(n+1))/2);
with
s+=c-((n*(n-1))>>1);
I dont have access to validate all inputs. and time range. but this one runs O(N) for sure. and have improved. run and let me know your feedback.i will provide details if necessary
public static int solution(int[]A,int K){
int minIndex=0;
int maxIndex=0;
int end=A.length;
int slize = end;
int startIndex = 0;
int diff = 0;
int minMaxIndexDiff = 0;
for(int currIndex=1;currIndex<end;currIndex++){
if(A[currIndex]>A[maxIndex]){
maxIndex=currIndex;
}else if(A[currIndex]<A[minIndex]){
minIndex=currIndex;
}
if( (A[maxIndex]-A[minIndex]) >K){
minMaxIndexDiff= currIndex- startIndex;
if (minMaxIndexDiff > 1){
slize+= ((minMaxIndexDiff*(minMaxIndexDiff-1)) >> 1);
if (diff > 0 ) {
slize = slize + (diff * minMaxIndexDiff);
}
}
if (minIndex == currIndex){
diff = currIndex - (maxIndex + 1);
}else{
diff = currIndex - (minIndex + 1);
}
if (slize > 1000000000) {
return 1000000000;
}
minIndex = currIndex;
maxIndex = currIndex;
startIndex = currIndex;
}
}
if ( (startIndex +1) == end){
return slize;
}
if (slize > 1000000000) {
return 1000000000;
}
minMaxIndexDiff= end- startIndex;
if (minMaxIndexDiff > 1){
slize+= ((minMaxIndexDiff*(minMaxIndexDiff-1)) >> 1);
if (diff > 0 ) {
slize = slize + (diff * minMaxIndexDiff);
}
}
return slize;
}
Get rid of the System.out.println() in the for loop :) you will be amazed how much faster your calculation will be
Nested assignments, i. e. instead of
t=i-ml;
s+=(t+c)-((t*(t+1))/(short)2);
i=ml;
ml++;
mxl=ml;
something like
s+=((t=i-ml)+c);
s-=((t*(t+1))/(short)2);
i=ml;
mxl=++ml;
sometimes occurs in OpenJDK sources. It mainly results in replacing *load bytecode instructions with *dups. According to my experiments, it really gives a very little speedup, but it is ultra hadrcore, I don't recommend to write such code manually.
I would try the following and profile the code after each change to check if there is any gain in speed.
replace:
if(Math.abs(A[ml]-A[mxl])<=K)
by
int diff = A[ml]-A[mxl];
if(diff<=K && diff>=-K)
replace
/2
by
>>1
replace
ml++;
mxl=ml;
by
mxl=++ml;
Maybe avoid array access of the same element (internal boundary checks of java may take some time)
So staore at least A[i] in a local varianble.
I would create a C version first and see, how fast it can go with "direct access to the metal". Chances are, you are trying to optimize calculation which is already optimized to the limit.
I would try to elimnate this line if(Math.abs(A[ml]-A[mxl])<=
by a faster self calculated abs version, which is inlined, not a method call!
The cast to (short) does not help,
but try the right shift operator X >>1 instead x / 2
removing the System.out.println() can speed up by factor of 1000.
But be carefull otherwise your whole algorithm can be removed by the VM becasue you dont use it.
Old code:
for(int i=0;i<100000;i++){
System.out.println(mezmeriz4r(A,K));
}
New code:
int dummy = 0;
for(int i=0;i<100000;i++){
dummy = mezmeriz4r(A,K);
}
//Use dummy otherwise optimisation can remove mezmeriz4r
System.out.print("finished: " + dummy);
Assume that I have a field called price for the documents in Solr and I have that field faceted. I want to get the facets as ranges of values (eg: 0-100, 100-500, 500-1000, etc). How to do it?
I can specify the ranges beforehand, but I also want to know whether it is possible to calculate the ranges (say for 5 values) automatically based on the values in the documents?
To answer your first question, you can get facet ranges by using the the generic facet query support. Here's an example:
http://localhost:8983/solr/select?q=video&rows=0&facet=true&facet.query=price:[*+TO+500]&facet.query=price:[500+TO+*]
As for your second question (automatically suggesting facet ranges), that's not yet implemented. Some argue that this kind of querying would be best implemented on your application rather that letting Solr "guess" the best facet ranges.
Here are some discussions on the topic:
(Archived) https://web.archive.org/web/20100416235126/http://old.nabble.com/Re:-faceted-browsing-p3753053.html
(Archived) https://web.archive.org/web/20090430160232/http://www.nabble.com/Re:-Sorting-p6803791.html
(Archived) https://web.archive.org/web/20090504020754/http://www.nabble.com/Dynamically-calculated-range-facet-td11314725.html
I have worked out how to calculate sensible dynamic facets for product price ranges. The solution involves some pre-processing of documents and some post-processing of the query results, but it requires only one query to Solr, and should even work on old version of Solr like 1.4.
Round up prices before submission
First, before submitting the document, round up the the price to the nearest "nice round facet boundary" and store it in a "rounded_price" field. Users like their facets to look like "250-500" not "247-483", and rounding also means you get back hundreds of price facets not millions of them. With some effort the following code can be generalised to round nicely at any price scale:
public static decimal RoundPrice(decimal price)
{
if (price < 25)
return Math.Ceiling(price);
else if (price < 100)
return Math.Ceiling(price / 5) * 5;
else if (price < 250)
return Math.Ceiling(price / 10) * 10;
else if (price < 1000)
return Math.Ceiling(price / 25) * 25;
else if (price < 2500)
return Math.Ceiling(price / 100) * 100;
else if (price < 10000)
return Math.Ceiling(price / 250) * 250;
else if (price < 25000)
return Math.Ceiling(price / 1000) * 1000;
else if (price < 100000)
return Math.Ceiling(price / 2500) * 2500;
else
return Math.Ceiling(price / 5000) * 5000;
}
Permissible prices go 1,2,3,...,24,25,30,35,...,95,100,110,...,240,250,275,300,325,...,975,1000 and so forth.
Get all facets on rounded prices
Second, when submitting the query, request all facets on rounded prices sorted by price: facet.field=rounded_price. Thanks to the rounding, you'll get at most a few hundred facets back.
Combine adjacent facets into larger facets
Third, after you have the results, the user wants see only 3 to 7 facets, not hundreds of facets. So, combine adjacent facets into a few large facets (called "segments") trying to get a roughly equal number of documents in each segment. The following rather more complicated code does this, returning tuples of (start, end, count) suitable for performing range queries. The counts returned will be correct provided prices were been rounded up to the nearest boundary:
public static List<Tuple<string, string, int>> CombinePriceFacets(int nSegments, ICollection<KeyValuePair<string, int>> prices)
{
var ranges = new List<Tuple<string, string, int>>();
int productCount = prices.Sum(p => p.Value);
int productsRemaining = productCount;
if (nSegments < 2)
return ranges;
int segmentSize = productCount / nSegments;
string start = "*";
string end = "0";
int count = 0;
int totalCount = 0;
int segmentIdx = 1;
foreach (KeyValuePair<string, int> price in prices)
{
end = price.Key;
count += price.Value;
totalCount += price.Value;
productsRemaining -= price.Value;
if (totalCount >= segmentSize * segmentIdx)
{
ranges.Add(new Tuple<string, string, int>(start, end, count));
start = end;
count = 0;
segmentIdx += 1;
}
if (segmentIdx == nSegments)
{
ranges.Add(new Tuple<string, string, int>(start, "*", count + productsRemaining));
break;
}
}
return ranges;
}
Filter results by selected facet
Fourth, suppose ("250","500",38) was one of the resulting segments. If the user selects "$250 to $500" as a filter, simply do a filter query fq=price:[250 TO 500]
There may well be a better Solr-specific answer, but I work with straight Lucene, and since you're not getting much traction I'll take a stab. There, I'd create a populate a Filter with a FilteredQuery wrapping the original Query. Then I'd get a FieldCache for the field of interest. Enumerate the hits in the filter's bitset, and for each hit, you get the value of the field from the field cache, and add it to a SortedSet. When you've got all of the hits, divide the size of the set into the number of ranges you want (five to seven is a good number according the user interface guys), and rather than a single-valued constraint, your facets will be a range query with the lower and upper bounds of each of those subsets.
I'd recommend using some special-case logic for a small number of values; obviously, if you only have four distinct values, it doesn't make sense to try and make 5 range refinements out of them. Below a certain threshold (say 3*your ideal number of ranges), you just show the facets normally rather than ranges.
You can use solr facet ranges
http://wiki.apache.org/solr/SimpleFacetParameters#Facet_by_Range